see [1]}, and we use it to eliminate the term \(J_{n - \frac{3}{2}}(z)\)  in the expression for \(\ddot{\xi _n}(z)\) . Then also \(J_{n - \frac{1}{2}}(z)\)  cancels out, such that we obtain the representation
\(\ddot{\xi _n}(z) = \sqrt{\textstyle {\frac{\pi }{2}}}\frac{\sqrt{z}}{z^2}\left(n(n + 1) - z^{2}\right)J_{n + \frac{1}{2}}(z) \) 
